Effects of harmonics on power systems - Part 1
Oct 1, 1999 12:00 PM, Sankaran, C.
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If not properly designed or rated, electrical equipment will often malfunction when harmonics are present in an electrical system.
Most people don't realize that harmonics have been around a long time. Since the first AC generator went online more than 100 years ago, electrical systems have experienced harmonics. The harmonics at that time were minor and had no detrimental effects.
Basic concept
A pure sinusoidal voltage is a conceptual quantity produced by an ideal AC generator built with finely distributed stator and field windings that operate in a uniform magnetic field. Since neither the winding distribution nor the magnetic field are uniform in a working AC machine, voltage waveform distortions are created, and the voltage-time relationship deviates from the pure sine function. The distortion at the point of generation is very small (about 1% to 2%), but nonetheless it exists. Because this is a deviation from a pure sine wave, the deviation is in the form of a periodic function, and by definition, the voltage distortion contains harmonics.
When a sinusoidal voltage is applied to a certain type of load, the current drawn by the load is proportional to the voltage and impedance and follows the envelope of the voltage waveform. These loads are referred to as linearloads (loads where the voltage and current follow one another without any distortion to their pure sine waves). Examples of linear loads are resistive heaters, incandescent lamps, and constant speed induction and synchronous motors.
In contrast, some loads cause the current to vary disproportionately with the voltage during each half cycle. These loads are classified as nonlinear loads, and the current and voltage have waveforms that are nonsinusoidal, containing distortions, whereby the 60-Hz waveform has numerous additional waveforms superimposed upon it, creating multiple frequencies within the normal 60-Hz sine wave. The multiple frequencies are harmonics of the fundamental frequency.
Normally, current distortions produce voltage distortions. However, when there is a stiff sinusoidal voltage source (when there is a low impedance path from the power source, which has sufficient capacity so that loads placed upon it will not effect the voltage), one need not be concerned about current distortions producing voltage distortions.
Examples of nonlinear loads are battery chargers, electronic ballasts, variable frequency drives, and switching mode power supplies. As nonlinear currents flow through a facility's electrical system and the distribution-transmission lines, additional voltage distortions are produced due to the impedance associated with the electrical network. Thus, as electrical power is generated, distributed, and utilized, voltage and current waveform distortions are produced.
Power systems designed to function at the fundamental frequency, which is 60-Hz in the United States, are prone to unsatisfactory operation and, at times, failure when subjected to voltages and currents that contain substantial harmonic frequency elements. Very often, the operation of electrical equipment may seem normal, but under a certain combination of conditions, the impact of harmonics is enhanced, with damaging results.
Motors
There is an increasing use of variable frequency drives (VFDs) that power electric motors. The voltages and currents emanating from a VFD that go to a motor are rich in harmonic frequency components. Voltage supplied to a motor sets up magnetic fields in the core, which create iron losses in the magnetic frame of the motor. Hysteresis and eddy current losses are part of iron losses that are produced in the core due to the alternating magnetic field. Hysteresis losses are proportional to frequency, and eddy current losses vary as the square of the frequency. Therefore, higher frequency voltage components produce additional losses in the core of AC motors, which in turn, increase the operating temperature of the core and the windings surrounding in the core. Application of non-sinusoidal voltages to motors results in harmonic current circulation in the windings of motors. The net rms current is [I.sub.rms] = [square root of [([I.sub.1]).sup.2] + [([I.sub.2]).sup.2] + [([I.sub.3]).sup.2] +] ..., where the subscripts 1, 2, 3, etc. represent the different harmonic currents. The [I.sub.2]R losses in the motor windings vary as the square of the rms current. Due to skin effect, actual losses would be slightly higher than calculated values. Stray motor losses, which include winding eddy current losses, high frequency rotor and stator surface losses, and tooth pulsation losses, also increase due to harmonic voltages and currents.
The phenomenon of torsional oscillation of the motor shaft due to harmonics is not clearly understood, and this condition is often disregarded by plant personnel. Torque in AC motors is produced by the interaction between the air gap magnetic field and the rotor-induced currents. When a motor is supplied non-sinusoidal voltages and currents, the air gap magnetic fields and the rotor currents contain harmonic frequency components.
The harmonics are grouped into positive (+), negative (-) and zero (0) sequence components. Positive sequence harmonics (harmonic numbers 1,4,7,10,13, etc.) produce magnetic fields and currents rotating in the same direction as the fundamental frequency harmonic. Negative sequence harmonics (harmonic numbers 2,5,8,11,14, etc.) develop magnetic fields and currents that rotate in a direction opposite to the positive frequency set. Zero sequence harmonics (harmonic numbers 3,9,15,21, etc.) do not develop usable torque, but produce additional losses in the machine. The interaction between the positive and negative sequence magnetic fields and currents produces torsional oscillations of the motor shaft. These oscillations result in shaft vibrations. If the frequency of oscillations coincides with the natural mechanical frequency of the shaft, the vibrations are amplified and severe damage to the motor shaft may occur. It is important that for large VFD motor installations, harmonic analyses be performed to determine the levels of harmonic distortions and assess their impact on the motor.
Transformers
The harmful effects of harmonic voltages and currents on transformer performance often go unnoticed until an actual failure occurs. In some instances, transformers that have operated satisfactorily for long periods have failed in a relatively short time when plant loads were changed or a facility's electrical system was reconfigured. Changes could include installation of variable frequency drives, electronic ballasts, power factor improvement capacitors, arc furnaces, and the addition or removal of large motors.
Application of nonsinusoidal excitation voltages to transformers increase the iron lesses in the magnetic core of the transformer in much the same way as in a motor. A more serious effect of harmonic loads served by transformers is due to an increase in winding eddy current losses. Eddy currents are circulating currents in the conductors induced by the sweeping action of the leakage magnetic field on the conductors. Eddy current concentrations are higher at the ends of the transformer windings due to the crowding effect of the leakage magnetic fields at the coil extremities. The eddy current losses increase as the square of the current in the conductor and the square of its frequency. The increase in transformer eddy current loss due to harmonics has a significant effect on the operating temperature of the transformer. Transformers that are required to supply power to nonlinear loads must be derated based on the percentages of harmonic components in the load current and the rated winding eddy current loss.
One method of determining the capability of transformers to handle harmonic loads is by k factor ratings. The k factor is equal to the sum of the square of the harmonic currents multiplied by the square of the frequencies.
k = [([I.sub.1]).sup.2]([1.sup.2]) + [([I.sub.2]).sup.2]([2.sup.2]) + [([I.sub.3]).sup.2]([3.sup.2]) + . . . + [([I.sub.n]).sup.2]([n.sup.2]).
where [I.sub.1] = ratio of fundamental current to total rms current, [I.sub.2] = ratio of second harmonic current to total rms current, [I.sub.3] = ratio of third harmonic current to total rms current, etc., and 1,2,3, ... n are harmonic frequency numbers. The total rms current is the square root of the sum of square of the individual currents.
By providing additional capacity (larger-size or multiple winding conductors), k factor rated transformers are capable of safely withstanding additional winding eddy current losses equal to k times the rated eddy current loss. Also, due to the additive nature of triplen harmonic (3, 9, 15, etc.) currents flowing in the neutral conductor, k rated transformers are provided with a neutral terminal that is sized at least twice as large as the phase terminals.
Example: A transformer is required to supply a nonlinear load comprised of 200A of fundamental (60 Hz), 30A of 3rd harmonic, 48A of 5th harmonic and 79A of 7th harmonic. Find the required k factor rating of the transformer:
Total rms current, I = [square root of [([I.sub.1]).sup.2] + [([I.sub.3]).sup.2] + [([I.sub.5]).sup.2] + [([I.sub.7]).sup.2]]
Total rms current, I = [square root of [(200).sup.2] + [(30).sup.2] + [(48).sup.2] + [(79).sup.2]] = 222.4A
[I.sub.1] = 200 / 222.4 = 0.899
[I.sub.3] = 30 / 222.4 = 0.135
[I.sub.5] = 48 / 222.4 = 0.216
[I.sub.7] = 79 / 222.4 = 0.355
k = [(0.899).sup.2][(1).sup.2] + [(0.135).sup.2] [(3).sup.2] + [(0.216).sup.2]([5).sup.2] + [(0.355).sup.2][(7).sup.2] = 8.31
To address the harmonic loading in this example, you should specify a transformer capable of supplying a minimum of 222.4A with a k rating of 9. Of course, it would be best to consider possible load growth and adjust the minimum capacity accordingly.
The photo (on page 33) shows one of the things that can happen when large nonlinear loads are present in a transformer. In this case, the nonlinear loads caused a substantial temperature rise. The unit had been installed to serve an online UPS source that produced high harmonic currents in the lines coming from the transformer. The darkened areas of the coils are due to the effect of heat caused by excess eddy current losses in the transformer's windings. Very often, the damage to the coils in a transformer is not known until a failure occurs.
Capacitor banks
Many industrial and commercial electrical systems have capacitors installed to offset the effect of low power factor. Most capacitors are designed to operate at a maximum of 110% of rated voltage and at 135% of their kvar ratings. In a power system characterized by large voltage or current harmonics, these limitations are frequently exceeded, resulting in capacitor bank failures. Since capacitive reactance is inversely proportional to frequency, unfiltered harmonic currents in the power system find their way into capacitor banks, These banks act like a sink, attracting harmonic currents, thereby becoming overloaded.
A more serious condition, with potential for substantial damage, occurs as a result of harmonic resonance. Resonant conditions are created when the inductive and capacitive reactances become equal in an electrical system. Resonance in a power system may be classified as series or parallel resonance, depending on the configuration of the resonance circuit. Series resonance produces voltage amplification and parallel resonance causes current multiplication within an electrical system. In a harmonic rich environment, both types of resonance are present. During resonant conditions, if the amplitude of the offending frequency is large, considerable damage to capacitor banks would result. And, there is a high probability that other electrical equipment on the system would also be damaged.
Fig. 1 (on page 36) shows a typical power system incorporating a distribution transformer ([T.sub.1]) and two variable frequency drives, each serving a 500hp induction motor. Assume that transformer [T.sub.1] is rated 3 MVA, 13.8kV-480V, 7.0% leakage reactance. With a 1000kvar capacitor bank installed on the 480V bus, the following calculations examine the power system for resonance. Where the secondary current of the 3MVA transformer is based at a potential of 480V, and neglecting utility source impedance, the transformer reactance at 7% results in an inductive reactance ([X.sub.L]) of 0.0161 ohms as determined from the following calculations, based upon a delta electrical configuration [ILLUSTRATION FOR FIGURE 2 AND 3 OMITTED]:
Transformer line current ([I.sub.L]) = [VA transformer rating] / [([square root of 3])([V.sub.L])]
([I.sub.L]) = [(3)[(10).sup.6]] / [([square root of 3])(480)] = 3608A
Note: impedance values are calculated using the actual winding current ([I.sub.w]) and winding voltage ([V.sub.w]).
[I.sub.w] = [I.sub.L] / [square root of 3 ] = 3608 / [square root of 3] = 2083A
Winding voltage ([V.sub.w]) = line voltage ([V.sub.L]) = 480V
Percent reactance (7%) = ([I.sub.w])([X.sub.L]) / ([V.sub.w])
Inductive reactance ([X.sub.L]) = (.07)([V.sub.w]) / ([I.sub.w]) = (.07)(480) / (2083) [X.sub.L] = 0.0161 ohms
Inductance (L) = [X.sub.L] / 2[Pi]f = 0.0161 / (2)(3.14)(60) = (0.428)[(10).sup.-4] henry
For a delta connected capacitor, the following calculations are applicable:
Line current to capacitor bank ([I.sub.L]) = (capacity in var) / ([square root of 3])([V.sub.L]) [I.sub.L] = (1000)[(10).sup.3] / ([square root of 3])(480) = 1203A
Capacitor current ([I.sub.c]) = [I.sub.L] / [square root of 3] = 1203 / 1.732 = 694.6A
Capacitive reactance ([X.sub.c]) = [V.sub.L] / [I.sub.c] = 480 / 694.4 = 0.691 ohm Capacitance (C) = 1 / 2[Pi]f[X.sub.c] = 1 / (2)(3.14)(60)(0.691)= (38.4)[(10).sup.-4] farad
Resonance frequency ([f.sub.R]) = 1 / 2[Pi][square root of (L)(C)]
([f.sub.R])= 1 / (2)(3.14) [[square root of (0.428)[(10).sup.-4] (38.4)[(10.)sup. -4]]]
([f.sub.R]) = 1 / (6.28) [[square root of (0.428)(38.4)[(10).sup.-8]]] = 393 Hz
A different derivation must be carried out when using a wye-connected transformer and a wye-connected capacitor bank. The wye-connected arrangement is the one normally used when a secondary neutral is required. The following equations are applicable for wye configurations ([ILLUSTRATION FOR FIGURE 4 AND 5 OMITTED], on page 40):
For the transformer:
Transformer winding voltage ([V.sub.w]) = line voltage ([V.sub.L]) / [square root of 3] = 480 / [square root of 3] = 277V
Winding current ([I.sub.w]) = transformer capacity (VA) / ([V.sub.L])([square root of 3])
[I.sub.w] = (3)[(10).sup.6] / (480)([square root of 3])= 3608A
Inductive reactance ([X.sub.L]) = (.07)([V.sub.w]) / ([I.sub.w]) = (.07)(277) / (3608)
[X.sub.L] = 0.00537 ohms
Inductance (L) = [X.sub.L] / 2[Pi]f = 0.00537 / (2)(3.14)(60) = (14.3)[(10).sup.-6] henry
For the capacitor bank:
Capacitor bank current flow ([I.sub.c]) = (capacity in var) / ([square root of 3])([V.sub.L])
[I.sub.c] = (1000)[(10).sup.3] / ([square root of 3])(480) = 1203A
Capacitor voltage ([V.sub.c]) = line voltage ([V.sub.L]) / [square root of 3] = 480 / [square root of 3] = 277V
Capacitive reactance ([X.sub.c]) = [V.sub.c] / [I.sub.c] = 277 / 1203 = 0.23 ohm
Capacitance (C) = 1 / 2[Pi]f[X.sub.c] = 1 / (2)(3.14)(60)(0.23) = 0.0115 farad
Resonance frequency ([f.sub.R]) = 1 / 2[Pi][square root of (L)(C)]
([f.sub.R]) = 1 / (2)(3.14)[[square root of (14.3)[(10).sup.-6]] (0.0115)]
([f.sub.R]) = 1 / (6.28)[[square root of (0.16445)[(10).sup.-6]]] = 393 Hz
Note that the resonance frequency remains the same, whether for a delta-type circuit or for a wye-type circuit. However, this situation would change should the transformer be one type circuit and the capacitor another type circuit.
The system would therefore be in resonance at a frequency corresponding to the 6.6th harmonic (393/60 = 6.55). This is dangerously close to the 7th harmonic voltage and current produced in variable frequency drives.
The two 500-hp drives draw a combined line current of 1100A (a typical value assuming motor efficiency of 90% and a .9PF). If the current of the 7th harmonic component is assumed to be 1/7 of the fundamental current (typical in drive applications), then [I.sub.7] = 1100 / 7 = 157A. If the source resistance (R) for the transformer and the conductors causes a 1.2% voltage drop based on a 3MVA load flow, then R = (0.92)([10.sup.-3]) ohms. This is because the determination of the inductive reactance ([X.sub.L]) for the wye-connected transformer was 0.00537 ohms. Thus, R = (0.00537)(1.2%) / 7% (transformer leakage reactance) = (0.92)([10.sup.-3]) ohms.
The "Q" or "quality factor" of an electrical system is a measure of the energy stored in the capacitors and inductors in the system. The current amplification factor (CAF) in a parallel resonant circuit (such as where a transformer and a capacitor are in a parallel configuration) is approximately equal to Q. Actually, Q= (2)([Pi]) (maximum energy storage) / (energy dissipation/cycle) as follows:
Q = [(2)([Pi])][(1/2)(L)[([I.sub.M]).sup.2] / [(I).sup.2] (R/f)]
where [I.sub.M] (maximum current) = ([square root of 2])(I), thus,
Q = (2)([Pi])(f)(L) / R = [X.sub.L] / R
where CAF can be considerd Q or [X.sub.L] / R.
For the example, with the two 500-hp drives, CAF equals (7)([X.sub.L]) / R, where 7 is a multiplication factor representing the 7th harmonic (or 7 times the fundamental 60Hz); [X.sub.L] is the reactive impedance at 0.00537; and R = (0.92)([10.sup.-3]) ohms. Thus:
CAF = (7)(.00537) / (0.92)([10.sup.-3]) = 40.86
The resonant current ([I.sub.R]) equals (CAF)([I.sub.7]) = (40.86)(157A)= 6415A. This current circulates between the source and the capacitor bank. The net current in the capacitor bank ([I.sub.Q] is equal to 6527A, which is derived as follows:
([I.sub.Q]) = [square root of [([I.sub.R]).sup.2] + [([I.sub.C]).sup.2]] = [square root of [(6415).sup.2] + [(1203).sup.2]] = 6527A
The value of [I.sub.Q] will seriously overload the capacitors. If the protective device does not operate to protect the capacitor bank, serious damage will occur.
The transformer and the capacitor bank may also form a series resonance circuit and cause large voltage distortions and overvoltage conditions at the 480V bus. Prior to installation of a power factor improvement capacitor bank, a harmonic analysis must be performed to ensure that resonance frequencies do not coincide with prominent harmonic components contained in the voltages and currents.
Cables
The flow of normal 60-Hz current in a cable produces [I.sup.2]R losses and current distortion introduces additional losses in the conductor. Also, the effective resistance of the cable increases with frequency due to skin effect, where unequal flux linkages across the cross section of the cable causes the AC current to flow on the outer periphery of the conductor. The higher the frequency of the AC current, the greater this tendency. Because of both the fundamental and the harmonic currents that can flow in a conductor, it is important to make sure a cable is rated for the proper current flow.
A set of calculations should be carried out to determine a cable's ampacity level. To do so, the first thing is to evaluate the skin effect. Skin depth relates to the penetration of the current in a conductor and varies inversely as the square root of the frequency, as follows:
Skin depth ([Delta]) = S / [square root of f]
where "S" is a proportionality constant based on the physical characteristics of the conductor and its magnetic permeability and "f" is the frequency.
If [R.sub.dc] is the DC resistance of a conductor, the AC resistance ([R.sub.f]) at frequency "f" is given by the expression,
[R.sub.f] = (K)([R.sub.dc])
The value of K is determined from the table shown on page 42. Its value corresponds to the calculated value of the skin effect resistance parameter (X), where X can be calculated as follows:
X = 0.0636 [square root of f[Mu] / [R.sub.dc]]
For this calculation, 0.0636 is a constant for copper conductors, "f" is the frequency, [R.sub.dc] is the DC resistance per mile of the conductor, and [Mu] is the permeability of the conducting material. The permeability for nonmagnetic materials, such as copper, is approximately equal to 1 and this is the value used. Tables or graphs that contain values of X and K are normally available from conductor manufacturers. The value of K is a multiplying factor that is to be multiplied by the normal cable resistance.
Example: Find the 60-Hz and 300-Hz AC resistances of a 4/0 copper conductor that has a DC resistance ([R.sub.dc]) of 0.276 ohm per mile. Using the following equation
X = 0.0636[square root of f[Mu] /[R.sub.DC]] We find that [X.sub.60] = (.0636)[[square root of (60)(1) / .276]] = 0.938. And, the value of K from the table, when [X.sub.60] = 0.938, is approximately 1.004. Thus, the conductor resistance per mile at 60 Hz = (1.004)(0.276) = 0.277 ohm.
For 300 Hz, [X.sub.300] = (.0636) [[square root of (300)(1) / .276]] = 2.097. For this condition, the value of K, based on [X.sub.300] = 2.097 from the table, is approximately 1.092. And, the conductor resistance per mile at 300 Hz = (1.092)(0.276) = 0.301 ohm.
The ratio of resistance, which is also called the skin effect ratio (E), based on the 300 Hz resistance to the 60 Hz resistance = .301 / .277 = 1.09. As can be seen; E = [X.sub.n] / [X.sub.60]
A conservative expression for the current rating factor (q) for cables that carry harmonic currents is derived by adding the [I.sup.2]R losses produced by each harmonic frequency current component at the equivalent 60 Hz level, as follows:
q = [[I.sub.[1.sup.2]][E.sub.1] + [I.sub.[2.sup.2]][E.sub.2] + [I.sub.[3.sup.2]][E.sub.3] + ... [I.sub.[n.sup.2][E.sub.N] where [I.sub.1], [I.sub.2], [I.sub.3] ... [I.sub.n] are the ratios of the harmonic currents to the fundamental frequency current and [E.sub.1], [E.sub.2], [E.sub.3], ... [E.sub.E] are skin effect ratios. (ratio of the effective resistance of the cable at the harmonic frequency to the resistance at the fundamental frequency).
Example: Determine the current rating factor (q) for a 60-Hz cable required to carry a nonlinear load with the following harmonic characteristics: fundamental current = 190A, 5th harmonic current = 50A, 7th harmonic current = 40A, 11th harmonic current = 15A and the 13th harmonic current = 10A.
The skin effect ratios are as follows:
[E.sub.1] = 1.0; [E.sub.5] = 1.09; [E.sub.7] = 1.17; [E.sub.11] = 1.35; [E.sub.13] = 1.44.
As previously mentioned, the skin effect ratio (E), also called the ratio of resistance, equals [X.sub.n] / [X.sub.60]. As an example, the skin effect ratio for E5 is based on the ratio of the 300 Hz resistance to the 60Hz resistance, which is 0.301 / 0.277 = 1.09.
The harmonic current ratios are as follows:
[I.sub.1] = 190/190 = 1.0 [I.sub.5] = 50/190 = 0.263 [I.sub.7] = 40/190 = 0.210 [I.sub.11] = 15/190 = 0.079 [I.sub.13] = 10/190 = 0.053 q = [(1.0).sup.2](1.0) + [(0.263).sup.2](1.09) + [(0.210).sup.2](1.17) + [(0.079).sup.2](1.35) + [(0.053).sup.2](1.44)
q = 1.14
Because the cable must be able to handle both the fundamental and the harmonic loads, based upon the q factor, the cable must be rated for a minimum current of (1.14)(190) = 217A at 60 Hz.
TERMS TO KNOW
Eddy current losses: Power dissipated due to current circulating in metallic material (core, windings, case, and associated hardware in motors, etc.) as a result of electromotive forces induced by variation of magnetic flux.
Hysteresis: The energy loss in magnet material that results from an alternating magnetic field as the elementary magnets within the material seek to align themselves with the reversing magnetic field.
Impedance: The total opposition that an electric circuit presents to an alternating current. It is the measure of the complex resistive and reactive attributes of a component (conductor, machinery, etc.) or of the total system within an AC circuit. Impedance causes electrical loss and is usually manifested in the form of heat.
Iron losses: These consist of hysteresis and eddy current losses associated with the metal laminations in motors and generators.
C. Sankaran is Senior Engineer, Electro-Test, Inc., Renton, Wash
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Effects of harmonics on power systems - Part 2
Feb 1, 1999 12:00 PM, Sankaran, C.
4 Comments | Related Content ShareThis27
Once you've recognized that harmonics are in a circuit or in an electrical system, the next step is to carry out tests to determine the magnitude and type of harmonics.
As more and more VFDs, electronic ballasts, battery chargers, and static var compensators are installed in facilities, the problems related to harmonics are expected to get worse. As such, it's important that you' re able to determine harmonic levels and analyze system data so that you can implement corrective measures and avoid serious problems.
Extent of harmonics
While harmonic voltages and currents are, by themselves, imperceptible, the physical phenomena that accompany them are perceivable. The adverse effects of harmonics in electrical power systems are very real, and failures related to voltage and current harmonics very often occur without warnings. When you have an indication that harmonics are present, your next step is to carry out testing to measure their level in the power system; you'll need this information to determine what mitigation system to use.
The degree to which harmonics affect electrical power system components depends on several factors: physical location, installation practices, electrical loading, and ambient temperatures. This means the same magnitude of harmonics might affect two separate installations differently. Some of the symptoms associated with large magnitudes of harmonics include large neutral currents, excessive temperature rise, vibration, audible noise, and protective device malfunction.
If you think large harmonic components are present, you first should learn what the electrical system is actually carrying on its lines. To do this, you must use harmonic measurement meters. The following instrumentation are commonly employed.
True-rms meters
The root mean square (rms) value, also known as the effective value, is the true measure of electrical parameters. For example, rms current represents the net heating effect of current on electrical equipment, thereby determining the thermal rating of the equipment. Operation of fuses and thermal magnetic circuit breakers is based on rms current. In transformers, the rms voltages determine the magnetic flux density levels in the transformer core. The rms voltage ratings determine the operating limits of electrical equipment. The relationship between the rms, average, and peak values of a pure sinusoidal current waveform follows. Form factor (FF) and peak factor (PF) are two elements that further define electrical waveforms. For a pure sinusoidal wave, the following relationships are true:
[I.sub.AVE] = (2/[Pi]([I.sub.M]) = (0.636)([I.sub.M])
[I.sub.RMS] = (1/[square root of 2])([I.sub.M]) = (0.707)([I.sub.M])
FF = [I.sub.RMS] [divided by] [I.sub.AVE] = 1.11
PF = [I.sub.M] [divided by] [I.sub.RMS] = 1.414
where [I.sub.M] = peak current
[I.sub.RMS] = rms current
[I.sub.AVE] = average current
Harmonics in electrical systems distort the waveforms and alter the rms and average values. Under such conditions, the relationship between the rms, peak, and average values are not represented by the above equations.
Conventional analog style meters do not accurately measure the rms values of nonsinusoidal voltages and currents due to deficiencies in their response to higher frequency components. Some earlier forms of digital meters measure the average or peak values and use multiplication factors to derive rms values. In a harmonic-rich environment, this is not valid. True-rms meters, by a process that involves high rate of signal sampling, recreate the waveform, and use frequency transformation techniques to obtain the true-rms values. True-rms meters may indicate that harmonics are present in an electrical system but may not provide a breakdown of the significant harmonics.
Harmonic analyzers
Harmonic analyzers are effective instruments for determining the waveshapes of voltage and current and measuring the respective frequency spectrum. Several types of harmonic measuring instruments are available, with each type having a different capability.
The simplest ones measure single-phase harmonic voltage and current, and provide information on the harmonic spectrums. These handheld instruments are easy to carry around. Fig. 1 shows voltage and current waveshapes and their harmonic frequency distribution recorded using a handheld harmonic analyzer. In addition, power factor and phase angle information are also measured by the harmonic analyzer used. The data shown in developing Fig. 1 were measured at the supply terminals of a power distribution panel feeding main frame computer-type loads.
Three-phase harmonic analyzers measure the harmonic characteristics of the three phases and the neutral simultaneously. Furthermore, some of the 3-phase analyzers provide graphs of the current and voltage distortion variations with time. These graphs are useful for determining if adverse harmonic loading conditions exist within the facility during plant operation. Fig. 2, on page 62, shows voltage and current harmonic distortion measured by a 3-phase analyzer, with the instrument leads connected at a main switchboard supplying an office building.
In addition to harmonic measurement, some analyzers are capable of measuring power, power factor, and transient disturbance data to help assess power quality within the power system. As expected, these instruments are less portable and considerably more expensive than the simple handheld-type units.
Harmonic analyzers calculate the total harmonic distortion (THD) of the waveform, so that overall distortion limits, as established under the guidelines of professional organizations, such as the Institute of Electrical and Electronic Engineers (IEEE), or the International Electrotechnical Commission (IEC), are not exceeded. IEEE defines THD by the following equation:
THD = [square root of [[([I.sub.2]).sup.2] + [([I.sub.3]).sup.2] + [([I.sub.4]).sup.2] + [([I.sub.5]).sup.2] + ...[([I.sub.n]).sup.2]] [divided by] [I.sub.1]
where, [I.sub.1] is the fundamental component of the current (60 Hz in the U.S.) and [I.sub.2], [I.sub.3], [I.sub.4]...[I.sub.n] are the harmonic frequency components of the current (multiples of the fundamental frequency).
While using a harmonic analyzer, it's important that you verify that voltage and current transformers (PTs and CTs) used with the analyzer have satisfactory higher frequency response characteristics. Normally, these instrument transformers are designed for optimum performance up to a cutoff frequency, beyond which their accuracies drop off considerably, introducing errors in the measurement. For example, to measure waveform distortion data up to the 50th harmonic, the PTs and CTs must have a frequency response of at least 3 kHz. (50th harmonic x 60 Hz = 3000 Hz). The manufacturers of harmonic analyzers are usually able to provide CTs with excellent harmonic frequency response.
Using an oscilloscope
Oscilloscopes have traditionally been used to troubleshoot electronic and electrical circuits. The older analog scopes had low high-frequency response characteristics and were very limited in their ability to perform waveform analyses. The digital storage-style scopes, which are now available, in addition to performing their traditional functions, have the capability to acquire and store signals and perform mathematical operations for determining frequency characteristics. These units collect and store the waveform data using voltage and current probes. This data then is downloaded into computers and synthesized to determine the waveform frequency characteristics.
Using harmonic analyzers
Harmonic analyzers are provided with voltage probes and current sensors. Some analyzers have seven to nine channels for simultaneous measurement of 3-phase voltages and currents and the neutral (each channel reading one parameter). There are other instruments available that offer less channels. If you have just a limited number of electrical parameters to be read simultaneously, the latter devices will work fine. Fig. 3 shows a typical installation of a harmonic analyzer to measure harmonics in a 3-phase, 4-wire lighting panel. If you want to make power and power factor measurements, make sure the polarities of the voltage probes and the current sensors are properly maintained; failing to do this will produce inaccurate results.
For high voltage (greater than 600V) and high current installations, PTs and CTs should be used. Again, make sure the high frequency response of these instrument transformers is adequate.
The location where an analyzer is to be installed depends upon the type of data required. If you suspect that a certain piece of equipment is generating harmonics, then the analyzer must be located in the lines feeding the equipment at a location close to the equipment. You must understand that as you move the analyzer upstream toward the power source, the harmonic currents become a decreasing percentage of the total load. This is partly due to the combination of harmonic non-linear loads and linear loads at upstream locations. Also, harmonic component currents generated by the different sources have varying phase angles between them. The net effect is cancellation of some of the harmonic currents as the measurement location is moved upstream.
When taking measurements, remember that while certain combinations of operating conditions could subject an electrical system to dangerous levels of harmonic distortions, operating conditions may be normal during the rest of the time. The best way you can address this is to take readings for 24-hr or longer periods.
Analyses of test results
By examining harmonic voltage and current data, you can get important information about the operating characteristics of a power system. Abnormal voltage as well as abnormal current conditions can cause problems. For example, most kinds of electrical equipment have maximum voltage distortion ratings for satisfactory operation. When the voltage distortion exceeds the established tolerance, certain types of equipment can malfunction or fail.
For current distortion, the magnitude of the fundamental current and/or the frequency distribution and magnitude of the harmonic current can cause equipment failure. A by-product of current distortion is excess thermal stress, which is a leading cause of equipment failure.
From harmonic spectrum data, k factors can be calculated to see if transformers can safely handle the harmonic load currents.
Emergency engine-generator sets, installed to provide power during utility outages, usually aren't very large. Thus, they're very limited in their capacity to handle harmonic loads and may fail during an emergency.
Motors can experience mechanical failures due to shaft torsional oscillations produced by the flow of harmonic currents in the motor windings. Therefore, it is vital that you carefully analyze harmonic data so that measures can be taken to prevent serious damage to equipment. Of course, you have to obtain information on equipment ratings to make such judgments.
Waveform distortion signatures
Harmonic distortions are characterized by the nature of the source responsible for the distortion. By examining the waveform, it's possible that you can determine the nature of the load producing the distortion. For example, variable frequency drives (VFDs), which use bridge rectifier circuits, produce a unique current waveform with two humps. Computer loads produce sharp peaks due to capacitive charging currents drawn by the power supply. Fluorescent lighting currents exhibit a flat current waveform due to striking of the arc in the light bulbs; at this point, the voltage and the current across the arc become flat. Current drawn by large arc furnaces produces extreme waveform distortions, with unequal positive and negative half cycles of currents.
Certain types of equipment produce even-order harmonics. These harmonics (2nd, 4th, 6th, 8th, etc.) are insignificant if the current's positive and negative half cycles are equal, as in a symmetrical power system. Even-order harmonic frequency current is a product of dissimilar current draw during two half cycles. Metering equipment will not read even-order harmonics because these harmonics cancel [TABULAR DATA OMITTED] themselves out. However, when the current's positive and negative halves are not equal, and even-order harmonics are present, then metering equipment will measure what's going on.
Even-order harmonics, where the current's positive and negative halves are not equal, are produced by arc furnaces, single-phase bridge, and half-wave rectifier circuits (as used in battery charges and power supplies for plating operations), and by transformer magnetizing currents. When metering equipment measure conditions of even-order harmonics, there usually should be no cause for alarm because the equipment operates that way.
If silicon controlled rectifiers (SCRs) are on the line (as used in some VFDs), however, then a reading of even-order harmonics is an indication of malfunctioning, such as the SCRs being unmatched due to manufacturing imperfections. In this condition, the SCRs may not turn on or off precisely. Therefore, conduction timing is not equal, or the SCRs are firing incorrectly. In such instances, the current flow during the positive and negative half cycles occurs during different durations, resulting in current mismatch during the two half cycles.
A small amount of mismatch can be tolerated by the equipment, and the signals produced may not be significant enough to stand out when taking instrument readings. If the mismatch is extreme, instrumentation will readily show the even-order harmonics. Here, you should take prompt corrective action.
As you can see, it's important that you have an understanding of the types of equipment connected to the electrical system when taking measurements. To help you in identifying problem sources, some harmonic analyzer manufacturers have published books containing samples of signature waveforms. You can compare your waveform with those published to determine the problem source.
In some instances, taking harmonic measurement at one location is all you'll need to define the problem. In other cases, you'll have to do a complete harmonic survey and analysis to assess the harmonic problem. The harmonic survey might involve data collection at several locations using harmonic analyzers and meters as required. Once the data is collected, harmonic analysis must be performed to identify potential problems, such as series and parallel resonance, harmonic heating, and motor torsional oscillations.
IEEE 519-1992 compliance
The above analysis will also reveal if the facility's power system complies with requirements noted in IEEE 519-1992, Recommended Practice And Requirements For Harmonic Control In Electric Power Systems, as indicated in the table above, for harmonic current injection into the utility lines. These requirements have been established to ensure that excessive harmonic currents are not so injected, which would affect the quality of power to other users sharing the same power lines and further overstress utility equipment. Presently, a number of utilities are considering placing contractual limitations in their rate structure regarding harmonic injection by their customers. Noncompliance could lead to penalty charges, higher rate schedules, or even electric service cutoff.
The harmonic current limitations established by IEEE 519-1992 are also applicable to equipment within a facility, as implementation of the standard will help enhance good operation. The point of harmonic current measurement in this case is the common junction between the offending loads and other equipment. For an example of using the IEEE standard, and making reference to the table, assume [I.sub.SC]/[I.sub.L] is 16. Then the net harmonic current distortion of all the harmonics up to and including the 10th is not to exceed 4.0%; the net harmonic current distortion of all the currents 11th to 16th is not to exceed 2.0%; and so on. The THD due to all harmonics must not exceed 5.0%. The reasoning behind this form of graded limits is to ensure that the larger users supplied by the utility are not allowed to inject larger quantity of harmonic frequency currents than the smaller users. It's expected that the use of IEEE 519-1992 will result in satisfactory power system operation within a facility, without placing undue burden on other loads or other utility customers sharing the same power source.
Practicing safely is important
Personnel safety is of primary concern when installing harmonic measurement equipment in electrical circuits. Awareness of the dangers that exist in a situation is the first step toward personnel safety. This awareness must be augmented by education about proper safety procedures and about equipment needed to protect against each hazard. Obviously, one safe practice action is to deenergize the electrical equipment prior to the installation of any harmonic measuring instrumentation. A procedure for safely deenergizing an electrical circuit can involve two parts. The first step is to evaluate the circuit for switching points and possible back feeds. This is done by comparing the single-line diagram and other available information associated with the circuit being deenergized and then to prepare a plan for switching off the live connections. The second step is to perform the switching in the order established by the plan.
Once the circuit is deenergized, you must install locks and tags on all applicable disconnecting devices and handles to ensure that the circuit can't be energized. Lockout devices are now available for all sizes of switches, fuse clips, breakers, and other devices. If you can't use locks, a tag should be supplemented by at least one additional safety measure, such as racking out a drawout circuit breaker, or disconnecting load conductors.
After lockout is completed, you should verify that the circuit is, in fact, deenergized. This can be done by using test equipment rated for the system line-to-line voltage. Any test meter used to verify the circuit must be checked for proper operation before and after the measurements.
Sometimes, it's not practical to deenergize a circuit for installation of harmonic measuring instrumentation. In such cases, you should wear proper protective equipment when installing the instruments. This equipment includes fire-resistant clothing, safety glasses, safety hats, rubber mats, electrical gloves, and electrical sleeves. Also, a second person trained in CPR and other first aid should be present during the installation of test leads. Never attempt to install instrumentation test leads on energized high voltage circuits (above 480V). The photo above shows the proper method of installing probes in electrical equipment.
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Fundamentals of Harmonics
Jun 1, 1999 12:00 PM, By Ken Michaels, Bell South Corp.
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With the exception of the incandescent light bulb, every load today creates harmonics. Unfortunately, these loads vary with respect to their amount of harmonic content and response to problems caused by harmonics.
Harmonics: It surfaced as a buzzword in the early 1980s, making many people reconsider the effectiveness of their building's wiring system. Yet, many still view the concept as a relatively new phenomenon. However, harmonics have been around since well before the early '80s: The associated problems existed in the electrical world way back when transistor tubes were first used in the 1930s. Aside from grounding, many deem harmonics as one of the greatest concerns for the power quality industry today. In this issue, we'll discuss the fundamentals of harmonics and the problems it can cause within the premises wiring system.
What is harmonics? We define harmonics as voltages or currents at frequencies that are a multiple of the fundamental frequency. In most systems, the fundamental frequency is 60 Hz. Therefore, harmonic order is 120 Hz, 180 Hz, 240 Hz and so on. (For European countries with 50 Hz systems, the harmonic order is 100 Hz, 150 Hz, 200 Hz, etc.)
We usually specify these orders by their harmonic number or multiple of the fundamental frequency. For example, a harmonic with a frequency of 180 Hz is known as the third harmonic (60x3 = 180). In this case, for every cycle of the fundamental waveform, there are three complete cycles of the harmonic waveforms. The even multiples of the fundamental frequency are known as even-order harmonics while the odd multiples are known as the odd-order harmonics.
How do we create harmonics? Up until 1980, all loads were known as linear. This means if the voltage input to a piece of equipment is a sine wave, the resultant current waveform generated by the load is also a sine wave, as seen in Fig. 1 (in the original text).
In 1981, manufacturers of electronic hardware converted to an efficient type of internal power supply known as a switch-mode power supply (SMPS). The SMPS converts the applied voltage sine wave to a distorted current waveform that resembles alternating current pulses, as seen in Fig. 2 (in the original text). Obviously, the load doesn't exhibit a constant impedance throughout the applied AC voltage waveform.
Most utilization equipment today creates harmonics. In all likelihood, if a device converts AC power to DC power (or vice versa) as part of its steady-state operation, it's considered a harmonic current-generating device. These include uninterruptible power supplies, copiers, PCs, etc.
What are the effects of harmonics? The biggest problem with harmonics is voltage waveform distortion. You can calculate a relationship between the fundamental and distorted waveforms by finding the square root of the sum of the squares of all harmonics generated by a single load, and then dividing this number by the nominal 60 Hz waveform value. You do this by a mathematical calculation known as a Fast Fourier Transform (FFT) theorem. (FFT is beyond the scope of this article. IEEE's Standard Dictionary of Electrical and Electronic Terms gives a definition of Fourier series.) This calculation method determines the total harmonic distortion (THD) contained within a nonlinear current or voltage waveform.
Triplen harmonics.Electronic equipment generates more than one harmonic frequency. For example, computers generate 3rd, 9th, and 15th harmonics. These are known as triplen harmonics. They are of a greater concern to engineers and building designers because they do more than distort voltage waveforms. They can overheat the building wiring, cause nuisance tripping, overheat transformer units, and cause random end-user equipment failure.
Circuit overloading.Harmonics can cause overloading of conductors and transformers and overheating of utilization equipment, such as motors. Triplen harmonics can especially caus e overheating of neutral conductors on 3-phase, 4-wire systems. While the fundamental frequency and even harmonics cancel out in the neutral conductor, odd-order harmonics are additive. Even in a balanced load condition, neutral currents can reach magnitudes as high as 1.73 times the average phase current.
This additional loading creates more heat, which breaks down the insulation of the neutral conductor. In some cases, it can break down the insulation between windings of a transformer. In both cases, the result is a fire hazard. But, you can diminish this potential damage by using sound wiring practices.
When most electrical engineers design the building's wiring, they usually leave the sizing of the neutral conductor to the dictates of NEC. In most cases, the installed neutral is the same size as the phase conductors. However, the Notes to the Ampacity Tables (in NEC Art. 310) instruct you to consider the neutral conductor as a current-carrying conductor if electronic equipment or electronic ballasts are used at the site. This correlates into the neutral conductors being sized larger than they would be with conventional wiring means.
To be on the safe side, more engineers are doubling the size of the neutral conductor for feeder circuits to panelboards and branch circuit partition wiring to handle the additive harmonic currents.
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